Day 7: electric Fields
THE ELECTRIC FIELD
THE ELECTRIC FIELD
Until this week, most of the forces you studied resulted from the direct action or contact of one piece of
matter with another. From your direct observations of charged, metal-coated balls, it should be obvious
that charged objects can exert electrical forces on each other at a distance. How can this be? The
action at a distance that characterizes electrical forces, or for that matter gravitational forces, is in some
ways inconceivable to us. How can one charged object feel the presence of another and detect its
motion with only empty space in between? Since all atoms and molecules are thought to contain
electrical charges, physicists currently believe that all “contact” forces are actually electrical forces
involving small separations. So, even though forces acting at a distance seem inconceivable to most
people, physicists believe that all forces act at a distance.
Physicists now explain all forces between charged particles, including contact forces, in terms of
the transmission of traveling electromagnetic waves. We will engage in a preliminary consideration of
the electromagnetic wave theory toward the end of the semester. For the present, let’s consider the
attempts of Michael Faraday and others to explain action-at-a-distance forces back in the nineteenth
century. Understanding more about these attempts should help you develop some useful models to
describe the forces between charged objects in some situations.
To describe action at a distance, Michael Faraday introduced the notion of an electric
field emanating from a collection of charged objects and extending out into space. More
formally, the electric field due to a known collection of charged objects is represented by an

electric field vector at every point in space. Thus, the electric field vector, E , is defined as the

force, Fe , experienced by a very small positive test charge at a point in space divided by the
magnitude of the test charge qt. The electric field is in the direction of the force e on a small
positive “test” charge and has the magnitude of

 Fe
E
qt
where qt is the charge on a small test particle.
Gravitational field -- analogous to electric field:
If a helicopter drops you out of its door at some distance above the Earth's surface, the
gravitational force due to the huge mass of the Earth causes you to accelerate downward. We
say that the earth is the source of a gravitational field that exerts a downward force on your
body and causes it to accelerate. The earth's gravitational field and the gravitational force
exerted by the field is much less if you are very far from the earth's surface. The field and force
are also less if you are near the surface of the moon, which has somewhat less mass than the
earth.
There are four important ideas related to these thought experiments:
• a gravitational field is caused by the mass of some object, like the mass of the Earth or of the
Moon;
• the magnitude of the gravitational field depends on the magnitude of the mass that produces
the field;
• the magnitude of the gravitational field also depends on the distance from the mass source
causing the field; and
• a mass, such as your body, when placed in the gravitational field, experiences a gravitational
force.
HTTP://MEDIA.PEARSONCMG.COM/BC/AW_YOUNG_PHYSICS_11/PT2A/MEDIA/E
LECTRICITY/1104POINTCHARGE/MAIN.HTML
Question 1: Explore the electric field
Suppose that Ball 1 at the bottom of the simulation screen is the source of the electric field.
Increase the charge of Q2 to +4.0 x 10-8 C. Use it to probe the electric field produced by Q1. To
see this field, grab Q2 and move it around. Notice that if Q2 gets closer to the location of Q1, the
magnitude of the electric force exerted on Q2 increases--the electric field is greater. If Q2 is
moved farther away from Q1, the magnitude of the electric force exerted on Q2 decreases--the
electric field is less. After systematically moving Q2 around the region surrounding Q1,
summarize in words your observations about the direction of the electric field produced by Q1 at
different points in space and how the field's magnitude varies.
Question 2: Field due to single positive charge
Set Q1 = +10.0 x 10-8 C. Use Q2 = +4.0 x 10-8 C as the test charge to measure the electric field
due to source charge Q1. Move Q2 to a point 1.0 m to the right side of the source charge Q1
(when r12 = 100 cm--2.5 divisions on the screen). Observe the direction of the electric force
exerted by Q1 on Q2. Use the force shown in the simulation
to calculate the electric field caused by Q1 at that point.
Question 3: Representing an electric field
Vary the value and the sign of the charge that is the source of the field (use only one point charge
for this activity) and develop some rules for the way that the electric field is represented.
Consider in particular:
• the direction of the lines;
• where the lines start or end relative to the sign of the charge causing the field;
• the separation of the lines in a particular region relative to the magnitude of the field in that
region;
• the magnitude of the field on a line and in the dark region next to the line, and
• the number of field lines that emanate from or terminate on a point charge relative to the
magnitude and sign of that charge.
Question 4: Uniform Field
Explain why the word "uniform" was used in describing the electric field in the middle region
between the plates.
Question 5: Force on a charge in a uniform field
Adjust the plate charge per unit area to 0.6 x 10-8 C/m2 and the value of the charge q between the
plates to +0.5 x 10-8 C. Suppose you place +q in the middle near the top plate . Note the arrow
showing the force on q and the magnitude of that force. Is the force greater in magnitude, the
same, or less if q is moved directly below its present position so that it is near the bottom plate?
Justify your choice. After your prediction, move q down and compare the force on it when in this
new position.
Question 6: Force on a negative charge
If you leave q in the middle half way between the plates and change the sign of q from +0.5 x 108
C to -0.5 x 10-8 C, what happens to the force exerted by the field on q. After your prediction,
adjust the charge of q to check your thinking. Try different magnitudes of q and different signs.
Move q around to different places between the plates and observe the direction and magnitude of
the force. Then, move the charge to the sides of the plates where the electric field is not uniform.
Are your observations consistent with the equation that relates the field and the force:
.
SHOW ELECTRIC FIELD MODELS
ELECTRIC FIELD
DEFINITION:
Units: (SI: either N/C or V/m = Volts/meter)

E, the force per unit of charge, is determine at each point by the ratio of the force on an
inconsequentially small, positive charge qo divided by size of the charge.

The direction of E in space is the same direction that a small positive charge would move (from
rest) if placed at that location, i.e. the direction of the force.
Some properties of the E-field:

Surrounding a charge we assert that there is an Electric Field E that has an independent
existence.

The Electric Field at some point in space (the field point ) can be thought of as being detached
from the charge(s) (the source point(s) ) from which the Electric Field originates. Later you will
see that an electromagnetic wave - a light beam for example - is a detached electric and
magnetic fields moving through space together at the speed of light.

The Electric Field is a Vector Function
; meaning that - in general - the symbol
stands for not just one thing but an uncountable infinity of vectors, each with its own magnitude
and direction. A reference to
space.
is a reference to an infinite number of vectors at all points in
Force on a Charged Particle in an Electric Field:

Turning the definition of the electric field around it is easy to see that if a charge q is in an
electric field E then the force acting on the charge is:

The sign of q is important. If q is negative then the force will be in the opposite direction of the
E-field.

The advantage of using the concept of an E-field is that the force on a charge can be easily found
if you know the vector value of the E-field at the location of the charge even though the E-field
may arise from a complicate arrangement of a large number of charges.
Field Lines:
To help visualize the electric field one often constructs field lines.

Field lines begin on positive charges and end on negative charges. Field lines point out of
+charges and into -charges. (Field lines can start at or go off to infinity when there is only a
single charge.


The electric field at a given point is tangent to the field line passing through that point.
There are an infinite number of field lines surrounding a charge. Conventionally, the field lines
are displayed in such a way that the number of field lines is proportional to the magnitude of
the charge from which they originate or terminate.

In regions where two field lines get closer together, the magnitude of the electric field is
becoming larger. In regions where the E-field is becoming weaker, the field lines are spread out.
A diagram of the field lines surrounding charges is called a field map. A two-dimensional field
map is a cross-section of the actual three-dimensional structure.

Superposition Principle:

The Electric Field at a particular point in space (due to presence of a large number of charges) is
equal to the vector sum of the E-Fields due to each source charge.

The direction of the electric force on a negative charge is opposite to the direction of the Electric
field at the location of the charge.
Examples of the Generation of an Electric Field
Electric Field due to a Point Charge q
Magnitude:
Direction: Radially outwards for positive charges and radially inward for negative charges.
Derived from
Coulomb's Law

For a set of charges, the total E-fields is the vector sum of the E-field of each charge.
Electric Field from Two Point Charges Problem
A positive and a negative charge of the same magnitude are placed equal distances from the origin on
the x-axis. The negative charge -1.7uC is place a distance 11 cm to the right of the origin, and the
positive charge +1.7uC is placed a distance 11cm to the left of the origin.
(B) what is the size of the electric field at a distance of 3.80 mm above the origin on the yaxis?
Physical Process and Sketch:
Two opposite charges on the x-axis create a combined electric field surrounding them. What is the
electric field like along the y-axis ?
Relevant Physics:
The electric field at any point in space is the vector sum of all the electric fields at that point due all the
other charges in space. In this problem there are only two point charges each of which which creates an
electric field at every point in space. (The only exception is if the point is on top of one of the charges.
Then the electric field due to a charge at the location of the charge is infinite, and will then be
excluded.)
In general,
We have taken the absolute value of the negative charge so that E2 will have positive value. We
will deal with the direction of E2 separately. Technically, the magnitude of any vector quantity is
always a positive number.
One key to solving this type of problem is to sketch the location of the charges and the direction
of the E-fields created by the charges at the point in question. Remember that the E-field points
away from positive charges and towards negative charges. A sketch of the setup shows that r1and
r2 are equal as well as the magnitudes of 1 and 2.
Applying some trig to the triangles formed, we can determine the values of r1 or r2 (which we
will now call just r).
We can also determine the angle 1 or 2,
More importantly, we can find the x-components of E1 and E2 in terms of the parameters given,
It is obvious in hindsight that these two x-components are equal in magnitude since they are the
same distance from their respective charges. What is not so obvious is that they are both in the
same direction. If both charges were positive (or negative) then these two components would be
in the opposite directions.
The vertical components are also equal but in opposite directions. This happens because the Efield points away from positive charges and towards negative charges; thus E2,y is negative and
E1,y is positive along the y-axis. Consequently, they cancel each other out and there is no vertical
component of the E-field along the y-axis. (Note that if the two charges did not have the same
magnitude then there would be a residual y-component of the E-field) For completeness, the ycomponents of the E-Field are:
The direction of the electric field can get very confusing if you try to carry the sign of the charge
along with its magnitude. In this case you would have obtained a positive value for E2,y (unless
you made a second error of neglecting the sign of the angle).
(A) Find E(0,y) - an expression for the magnitude of the electric field on the y-axis as function of location
of the charges and their magnitude.
The direction of the electric field is easy to determine since the y-components of electric field of
the two charges cancel each other out. From the sketch we can see that the E-field will be
horizontal to the right, i.e. its angle is zero.
The E-field is just equal to the sum of the two horizontal components of the E-field due to the
two charges.
What we have found is E(0,y) for this arrangement of charges.
Observe that when y becomes very large, the expression approaches 1/ or zero. This is to be
expected since the net charge of the two charges would appear to merge from a great distance
and the net charge is zero.
At the origin, the E-field goes over to
The two E-fields add in between a positive and negative charge. Note that had you just been
trying to find the E-field at the origin and not dropped the sign on the negative sign you might
have gotten something like,
This is incorrect. In fact, the E-field at the origin will only be zero if the two charges had the
same sign not opposite signs.
(B) Find the strength of the E-field at y = .00380 m, a = .110 m, and q = 1.70x10-6 C.
Simple substitution of the results of part A yields,
This may seem like a large number but it is typical of the size of the E-field for static charges in
the microcolumb range. This size electric field is also typical of the E-field at the edges of cell
membranes. Also see Typical Values of Electric Field.
(C) Find y such that E(0,y) = 7.30x103 N/C.
The easiest way to find y would be to use the solve-mode on your calculator since we already
found the expression to solve in part A.
Using E instead of E(0,y), let us solve for E symbolically.
Of course -.765 m is also a location on the y-axis where the value of the E-field is 7.30x103 N/C.
(D) What is magnitude and direction of the force on a -8.60x10-5 C charge placed .0038 m above the yaxis.
Ones first impulse might be to calculate the Coulomb force that each of the two original charges
have on the new third charge and then add the forces vectorly. This is entirely unnecessary since
we know the E-field at the location of the third charge from part B. What we are about to do
represents one of the main advantages of introducing the concept of the electric field in the first
place. Namely, if one knows the size of the charge and the size of the electric field at the charge's
location, then one can find the force from,
For our problem,
The minus sign means that the direction of force is opposite that the electric field. In this case, it
means that the force is to left.
The large force results from the fact that a microcolumb of localized static charge is a large
charge in practical terms. The practical, upper-limit to which an ordinary-sized object can be
charged is about 10 nC/cm2. This means that a sphere charged with -8.60x10-5 C would have to
have a minimum radius of 26 cm.
SUPERPOSITION OF ELECTRIC FIELD VECTORS
The fact that electric fields from charged objects that are distributed at different locations act along a
line between the charged objects and the point in space of interest is known as linearity. The fact that
the vector fields due to charged objects at different points in space can be added together is known as
superposition. These two properties of the Coulomb force and the electric field that derives from it are
very useful in our endeavor to calculate the value of the electric fields due to a collection of point
charges at different locations. This can be done by finding the value of the E-field vector from each
point charge and then using the principle of superposition to determine the vector sum of these
individual electric field vectors.
Activity: Electric Field Vectors from Two Point Charges
a. Look up the equations for Coulomb’s law and the electric field from a point charge in your
textbook. Also check out the value of any constants you would need to calculate the actual value
of the electric field from a point charge. List the equations and any needed constants in the space
below.
b. Use a spreadsheet to calculate the magnitude of the electric field (in N/C) at distances of 0.5, 1.0,
–9
1.5 . . . 10.0 cm from a point charge of 2.0  10 C. Be careful to use the correct units (that is,
convert the distances to meters before doing the calculation).
c. The following graph shows two point particles with charges of
–9
–9
+2.0  10 N and –2.0  10 C that are separated by a distance of 8.0 cm. Use the principle of
linearity to draw the vector contribution of each of the point charges to the electric field at each of
the four points in space shown below. Use your spreadsheet results and a scale in which the
4
vector is 1 cm long for each electric field magnitude of 1.0  10 N/C. Then use the principle of
superposition and the polygon method to find the resultant E vector at each point. Hint: One of
the point charges will attract a positive test charge and the other will repel it.
Typical Values of the Electric Field
Source
Field Size
N/C or V/m
Electrical Wiring
10-2
Radio Waves
10-1
Class Room
10-100
Sunlight
103
Thunderstorm
104
Breakdown of Air
106
On an Electron in Hydrogen Atom
1011
Surface of Neutron Star
1014
THE ELECTRIC FIELD FROM AN EXTENDED CHARGE DISTRIBUTION
Although electric charges will not usually distribute themselves uniformly throughout a conductor,
charge can be distributed uniformly through an insulator. If charge is distributed uniformly throughout
a continuous extended insulated object, it can be divided into small segments each of which contains a
charge q. Then, by assuming that each segment behaves like a tiny point charge, the electric field at a
point P in space due to each segment can be calculated. The total electric field at P is simply the vector
sum of the contributions of each of the charge segments. This process yields an approximate value of
the electric field at point P. Such approximate values can be calculated quite readily using a computer
spreadsheet. To get a more exact value we must sum up infinitely many infinitesimally small elements
of charge q. This is what mathematical integration is all about.

The goal of this section of the activity guide is to calculate the electric field E corresponding to
a continuous charge distribution on a rod at two points in space, P and P', as shown below.
Diagram of a charged rod broken into ten imaginary “point” charges for a spreadsheet calculation of the
electric field at points P and P'.
Each of these calculations will be done two ways: (1) doing an approximate numerical calculation
with the spreadsheet, and (2) doing an “exact” integration. These two methods of calculation will be
compared with each other. You could extend the calculation to other points in space and graph the change
in field as a function of the distance from the rod along a line through the axis of the rod and along a line
perpendicular to the rod.
Activity: E-Field Vectors from a Uniformly Charged Rod

In each case, draw the magnitude and direction of ten vectors, Ei at point P or P'. Each vector

approximates the relative scale of each Ei at the two points (draw longer arrows for the vectors
corresponding to charge elements closer to P or P'). Use the following diagrams and draw a
resultant vector in each case.
a. Parallel to the axis of the rod
b. Perpendicular to the axis of the rod
Activity: Electric Field Calculations Along the Axis of a Rod
a. Consider a rod of length L that is divided into n segments. If the total charge on the rod is given
by Q, show that the charge q in each segment X of the rod is given by Q = (Q/L) X.
b. Use the spreadsheet to find the electric fields due to the ten elements numerically. You should

probably define three columns: Charge Element #, x, and E . Once the calculations are done you can


sum up the E ’s to get the value of E . Do this for both of the indicated points. Include a snapshot of
your spreadsheet for the Blog.
Electric Field Due To A Differential Charge
The general idea is to imagine that you replace charge distribution by an infinite number of
differentially small charges dq that fill the region where the charge is distributed. Then pick one
dq so that it is located at some representative location in the charge distribution.
Using this dq as a source, determine the E-field at the point where you want to know its value
If you could move the dq all around inside of the charge distribution and sum up all the
contributions from each dq, you would obtain the value of the E-field . You accomplish this task
by integrating dq over the charge distribution.
The magnitude of dq can be determined if you know (or can calculate) the type of charge density
involved , or .
Replacing dq by the appropriate charge density you can turn the intergrals over charge into a
spacial intergrals which can be integrated to find the components of the E-field. For a specific
example see calculus derivation of the E-field near a infinite line charge
c.Set up the integral for E and solve it to show that the equation relating the magnitude of the electric
field, E, is given by
E
ke Q
d L  d
Then substitute the values for L, d, and Q to find a numerical value of E. Hint: What are the
limits of integration—that is, what is the range of x in which the charged rod exists?
d. How do the numerical and “exact” values compare? Compute the percent discrepancy. How
could you make the numerical method more “exact”?
The first set of calculations for the E-field along the axis of the rod was relatively easy because all the
electric field vectors lie along a single line. Now you are to do a calculation of the E-field perpendicular
to the axis of the rod. In this case we have to consider both the x- and y-components of the electric field
resulting from the charge on each element.
Electric Field due to a Infinite Line Charge with Uniform Charge Density 
 = Charge per unit length (SI: C/m)
Magnitude:
Direction:
Derivation
Radially outwards for a positive  perpendicular to the
line charges.
It is assumed that the charges on the line are fixed in position. A conduction rod would produce
just an E-field, but as soon as you put any charge near the rod, the distribution of charge on the
rod would change.
Calculus Derivation:





Since the line is infinite we could place the origin under the point of interest and turn the
problem into two semi-infinite line charges that meet at the origin.
Next consider two infinitesimal, differential charges dq that are an equal distance from the
origin. The x-components of the E-field of the two differential charges are in opposite directions
at the any midway point and cancel each other out.
Tthe E-field at any distance from the wire will only point along the y-axis, perpendicular to the
wire. More generally, the will be radial outwards or inwards towards the wire.
The magnitude of the E-field can be found by determining the contributions from all dq's
betweem zero and infinity (i.e. by integrating the contribution of dq) and then doubling the
value, E = 2E y.
Treating the differential charge dq as a point charge, the y-component of the electric field is

The differential charge can be written as dq = dx. Where  is charge per unit length which is
constant.

Integrating between zero and infinity,

Using an integral table,

Note that as x approaches infinity the value of y can be neglected relative to x in the
denominator of the first term so that this term approaches x/x = 1.

We only solve the problem in one plane passing through the line charge - the xy-plane - but our
results are true for any plane and any point. Changing y to r to represent the distance of the
point from the line charge, we obtain the generic answer.

What is amazing is that the E-field is not infinite everywhere in space since there is an infinite
amount of charge on the infinite line charge. This happens because the E-field of a point charge
drops off as one over distance squared.
Activity: Electric Field Calculations  to the Axis
Explain why the x-component of the total E-vector at point P' should be zero. Hint: The
argument used to show this is known as a symmetry argument.
Then substitute the values for L, d, and Q to find a numerical value of E. Hint: What are the limits of
integration—that is, what is the range of x in which the charged rod exists?
d. How do the numerical and “exact” values compare? Compute the percent discrepancy. How
could you make the numerical method more “exact”
https://phet.colorado.edu/en/simulation/electric-hockey
Give screen shots of success on levels 1 and 2. What is the minimum number of charges you can use for
level 1 and level2?
Electric Field on the Axis of a Charged Ring with a Uniform Charge Density
Q = Total Charge on the Ring of radius R
Magnitude:
Direction: Outwards away from the ring along the axis of the ring if Q is positive.

The ring is assumed to lie in zy-plane with its axis along the x-axis

The E-field (as a function of its location) at points other than along the axis is very complex and it
can not easily be expressed by such a simple function.
Electric Field on the Axis of Charged Ring

This is a very limited solution to the more general problem of finding the E-field at any point in
space near a ring with a charge Q spread uniformly around the ring.

Along the axis (say the x-axis) the perpendicular components of the E-field due to charges
spread around the ring cancel each other out. There is just as much charge on one side of the
ring and the other.

The net E-field (on the axis) is along the axis, outwards from the ring if the charge Q is positive
and towards the ring in the charge Q is negative.

The charge density  on the ring is just the total charge Q on the ring divided by its
circumference 2 R.

Treating the differential charge dq as a point charge the differential electric field at a distance x
from the center of the ring (the origin) is given by

Since the sum of the y-components cancel out, the magnitude of the electric field is equal to the
sum of the x-components of the E-field. We can express this as an integral over the differential
arc length ds,
Note that x and r are both constant from any location on the ring to the point where we are calculating
the E-field.
There two interesting limits.
1. At the very center of the ring the E-field is zero at one would expect. The E-fields of the charges
spread around the ring cancel each other out.
2. When one get very far from the ring so that x >> R, then R can be neglected in the denominator
compared to x, and the loop looks like a point charge at large distances.

Electric Field due to Infinitely Flat, Charged-Plane with a Uniform Charge
Density
 = Charge per unit area on the plane (SI: C/m 2)
Magnitude:
Direction: Outwards perpendicular to the plane.

For an infinite sheet the E-field is the same at any distance from the sheet, i.e. the E-field is
constant.

This expression is useful for electrical conductors because the charge spreads out on the
conductor's surface forming a surface charge distribution . Although the charge density  is
not necessarily constant over the surface, but if you get close enough, the conductor's surface
will look like a infinitely charge sheet. The electric field will be perpendicular to the surface and
equal to half of a charged sheet because the E-field inside the conductor is zero.
E = /o